3.207 \(\int \frac{(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac{7}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=156 \[ \frac{4 a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]

[Out]

((-4 - 4*I)*a^(5/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + (4*a^2*Sqrt[
a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) - (((2*I)/3)*a*(a + I*a*Tan[c + d*x])^(3/2))/(d*Tan[c + d*x]^(3/
2)) - (2*(a + I*a*Tan[c + d*x])^(5/2))/(5*d*Tan[c + d*x]^(5/2))

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Rubi [A]  time = 0.281975, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3548, 3545, 3544, 205} \[ \frac{4 a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(7/2),x]

[Out]

((-4 - 4*I)*a^(5/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + (4*a^2*Sqrt[
a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) - (((2*I)/3)*a*(a + I*a*Tan[c + d*x])^(3/2))/(d*Tan[c + d*x]^(3/
2)) - (2*(a + I*a*Tan[c + d*x])^(5/2))/(5*d*Tan[c + d*x]^(5/2))

Rule 3548

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(d*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f*m*(c^2 + d^2)), x] + Dist[a/(a*c - b*d), Int[(a
+ b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*
d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n + 1, 0] &&  !LtQ[m, -1]

Rule 3545

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*b*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m - 1)*(a*c - b*d)), x] + Dist[(2*a^2)/(
a*c - b*d), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f},
x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n, 0] && GtQ[m, 1/2]

Rule 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac{7}{2}}(c+d x)} \, dx &=-\frac{2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)}+i \int \frac{(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-(2 a) \int \frac{(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{4 a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\left (4 i a^2\right ) \int \frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{4 a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)}-\frac{\left (8 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{-i a-2 a^2 x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}\\ &=-\frac{(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}+\frac{4 a^2 \sqrt{a+i a \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{2 i a (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac{3}{2}}(c+d x)}-\frac{2 (a+i a \tan (c+d x))^{5/2}}{5 d \tan ^{\frac{5}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 3.36801, size = 187, normalized size = 1.2 \[ -\frac{4 i a^2 e^{-i (c+d x)} \cot (c+d x) \left (e^{i (c+d x)} \left (-35 e^{2 i (c+d x)}+26 e^{4 i (c+d x)}+15\right )-15 \left (-1+e^{2 i (c+d x)}\right )^{5/2} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt{a+i a \tan (c+d x)}}{15 d \sqrt{-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}} \left (-1+e^{4 i (c+d x)}\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(7/2),x]

[Out]

(((-4*I)/15)*a^2*(E^(I*(c + d*x))*(15 - 35*E^((2*I)*(c + d*x)) + 26*E^((4*I)*(c + d*x))) - 15*(-1 + E^((2*I)*(
c + d*x)))^(5/2)*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*
x]])/(d*E^(I*(c + d*x))*Sqrt[((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x)))]*(-1 + E^((4*I)*(c +
d*x))))

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Maple [B]  time = 0.036, size = 412, normalized size = 2.6 \begin{align*}{\frac{{a}^{2}}{15\,d}\sqrt{a \left ( 1+i\tan \left ( dx+c \right ) \right ) } \left ( 15\,i\sqrt{ia}\sqrt{2}\ln \left ({\frac{1}{\tan \left ( dx+c \right ) +i} \left ( 2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) \right ) } \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{3}a+15\,\sqrt{ia}\sqrt{2}\ln \left ({\frac{2\,\sqrt{2}\sqrt{-ia}\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }-ia+3\,a\tan \left ( dx+c \right ) }{\tan \left ( dx+c \right ) +i}} \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{3}a+76\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\sqrt{ia} \left ( \tan \left ( dx+c \right ) \right ) ^{2}+60\,\ln \left ( 1/2\,{\frac{2\,ia\tan \left ( dx+c \right ) +2\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}+a}{\sqrt{ia}}} \right ) \sqrt{-ia} \left ( \tan \left ( dx+c \right ) \right ) ^{3}a-22\,i\tan \left ( dx+c \right ) \sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{ia}\sqrt{-ia}-6\,\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }\sqrt{-ia}\sqrt{ia} \right ) \left ( \tan \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}{\frac{1}{\sqrt{a\tan \left ( dx+c \right ) \left ( 1+i\tan \left ( dx+c \right ) \right ) }}}{\frac{1}{\sqrt{ia}}}{\frac{1}{\sqrt{-ia}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(7/2),x)

[Out]

1/15/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2/tan(d*x+c)^(5/2)*(15*I*(I*a)^(1/2)*2^(1/2)*ln((2*2^(1/2)*(-I*a)^(1/2)*(a
*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c)^3*a+15*(I*a)^(1/2)*2^(1/2)*
ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*tan(d*x+c
)^3*a+76*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*(I*a)^(1/2)*tan(d*x+c)^2+60*ln(1/2*(2*I*a*tan(d*x+
c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*tan(d*x+c)^3*a-22*I*tan(d*
x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)*(-I*a)^(1/2)-6*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-
I*a)^(1/2)*(I*a)^(1/2))/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(1/2)/(-I*a)^(1/2)

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Maxima [B]  time = 2.81366, size = 1725, normalized size = 11.06 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

-1/225*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*((-(900*I - 900)*a^2*cos(3*d*x
+ 3*c) + (930*I - 930)*a^2*cos(d*x + c) + (900*I + 900)*a^2*sin(3*d*x + 3*c) - (930*I + 930)*a^2*sin(d*x + c))
*cos(3/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + ((900*I + 900)*a^2*cos(3*d*x + 3*c) - (930*I + 93
0)*a^2*cos(d*x + c) + (900*I - 900)*a^2*sin(3*d*x + 3*c) - (930*I - 930)*a^2*sin(d*x + c))*sin(3/2*arctan2(sin
(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))*sqrt(a) + (((900*I - 900)*a^2*cos(2*d*x + 2*c)^2 + (900*I - 900)*a^2*s
in(2*d*x + 2*c)^2 - (1800*I - 1800)*a^2*cos(2*d*x + 2*c) + (900*I - 900)*a^2)*arctan2((cos(2*d*x + 2*c)^2 + si
n(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - c
os(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*
x + 2*c), -cos(2*d*x + 2*c) + 1)) - sin(d*x + c)) + (-(450*I + 450)*a^2*cos(2*d*x + 2*c)^2 - (450*I + 450)*a^2
*sin(2*d*x + 2*c)^2 + (900*I + 900)*a^2*cos(2*d*x + 2*c) - (450*I + 450)*a^2)*log(cos(d*x + c)^2 + sin(d*x + c
)^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(cos(1/2*arctan2(sin(2*d*x + 2*c)
, -cos(2*d*x + 2*c) + 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2) - 2*(cos(2*d*x + 2*
c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c)
+ 1))*sin(d*x + c) + cos(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))))*(cos(2*d*x + 2
*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + (((900*I - 900)*a^2*cos(5*d*x + 5*c) - (7
50*I - 750)*a^2*cos(3*d*x + 3*c) + (210*I - 210)*a^2*cos(d*x + c) - (900*I + 900)*a^2*sin(5*d*x + 5*c) + (750*
I + 750)*a^2*sin(3*d*x + 3*c) - (210*I + 210)*a^2*sin(d*x + c))*cos(5/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x +
2*c) + 1)) + ((-(240*I - 240)*a^2*cos(d*x + c) + (240*I + 240)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 - (240*I
- 240)*a^2*cos(d*x + c) + (-(240*I - 240)*a^2*cos(d*x + c) + (240*I + 240)*a^2*sin(d*x + c))*sin(2*d*x + 2*c)^
2 + (240*I + 240)*a^2*sin(d*x + c) + ((480*I - 480)*a^2*cos(d*x + c) - (480*I + 480)*a^2*sin(d*x + c))*cos(2*d
*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) + (-(900*I + 900)*a^2*cos(5*d*x + 5*c) +
(750*I + 750)*a^2*cos(3*d*x + 3*c) - (210*I + 210)*a^2*cos(d*x + c) - (900*I - 900)*a^2*sin(5*d*x + 5*c) + (75
0*I - 750)*a^2*sin(3*d*x + 3*c) - (210*I - 210)*a^2*sin(d*x + c))*sin(5/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x
+ 2*c) + 1)) + (((240*I + 240)*a^2*cos(d*x + c) + (240*I - 240)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (240*I
+ 240)*a^2*cos(d*x + c) + ((240*I + 240)*a^2*cos(d*x + c) + (240*I - 240)*a^2*sin(d*x + c))*sin(2*d*x + 2*c)^
2 + (240*I - 240)*a^2*sin(d*x + c) + (-(480*I + 480)*a^2*cos(d*x + c) - (480*I - 480)*a^2*sin(d*x + c))*cos(2*
d*x + 2*c))*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))*sqrt(a))/((cos(2*d*x + 2*c)^2 + sin(2*d
*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(5/4)*d)

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Fricas [B]  time = 2.42411, size = 1337, normalized size = 8.57 \begin{align*} \frac{\sqrt{2}{\left (208 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 72 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 160 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 120 i \, a^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + 15 \, \sqrt{\frac{32 i \, a^{5}}{d^{2}}}{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac{{\left (4 \, \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} + i \, \sqrt{\frac{32 i \, a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{2}}\right ) - 15 \, \sqrt{\frac{32 i \, a^{5}}{d^{2}}}{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac{{\left (4 \, \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - i \, \sqrt{\frac{32 i \, a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{2}}\right )}{30 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} - 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

1/30*(sqrt(2)*(208*I*a^2*e^(6*I*d*x + 6*I*c) - 72*I*a^2*e^(4*I*d*x + 4*I*c) - 160*I*a^2*e^(2*I*d*x + 2*I*c) +
120*I*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I
*d*x + I*c) + 15*sqrt(32*I*a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c)
- d)*log(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x
+ 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + I*sqrt(32*I*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*
I*d*x - 2*I*c)/a^2) - 15*sqrt(32*I*a^5/d^2)*(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x
+ 2*I*c) - d)*log(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^
(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - I*sqrt(32*I*a^5/d^2)*d*e^(2*I*d*x + 2*I*c)
)*e^(-2*I*d*x - 2*I*c)/a^2))/(d*e^(6*I*d*x + 6*I*c) - 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) - d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))**(5/2)/tan(d*x+c)**(7/2),x)

[Out]

Timed out

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Giac [A]  time = 1.37264, size = 193, normalized size = 1.24 \begin{align*} \frac{\left (i - 1\right ) \, \sqrt{-2 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 2 \, a^{2}}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{4} \log \left (\sqrt{i \, a \tan \left (d x + c\right ) + a}\right )}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} - 6 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a + 14 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} - 16 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} + 9 \,{\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} - 2 \, a^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

(I - 1)*sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(I*a*tan(d*x + c) + a)^2*a^4*log(sqrt(I*a*tan(d*x + c) + a))
/((I*a*tan(d*x + c) + a)^5 - 6*(I*a*tan(d*x + c) + a)^4*a + 14*(I*a*tan(d*x + c) + a)^3*a^2 - 16*(I*a*tan(d*x
+ c) + a)^2*a^3 + 9*(I*a*tan(d*x + c) + a)*a^4 - 2*a^5)